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anonymous
 5 years ago
Find all critical points of f(z)=(3z^26)^2(15z)^4
anonymous
 5 years ago
Find all critical points of f(z)=(3z^26)^2(15z)^4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Take the derivative of f(z) with respect to z and set it equal to zero. The values of z that result will be critical points; don't know if you want to distinguish between different types though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I understand how to find them  I was having problems with setting the derivative to zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright; give me a minute and I'll try it out. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've got f'(z)=6(z^22)(6z)(15z)^415(z^22)4(15z)^3 for the derivative. Look like what you got?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you differentiate, it should come out to this:\[f'(z) = 6*(15z)^4*(3z^26)*2z20*(15z)^3*(3x^26)^2\] Remember to not get sidetracked by careless errors; they're very prevalent in this problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, accidentally stuck an x into the last factor when it was meant to be a z :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The image cuts off at (3x^2  6)...is that the end? Just making sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I get a pretty simple quadratic equation in the end..yep.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm..How did you get to the quadratic equation? Sorry I don't understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just moved the entire right term (starting with 20*(15z)^3...) to the left hand side of the equation, and started simplifying. Things begin cancelling immediately and it breaks down into a quadratic equation (or so I believe)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You set them equal to each other and simplify? How can they simplify if they are being multiplied to each other  I though you can't do that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No...I took that derivative and set it equal to zero, then when I moved the term beginning with 20 to the opposite side (where the 0 would be) I began cancelling common factors.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you get 0 = 60z^2 +12z  20 at some point?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you get 0 = 60z^2 +12z  20 at some point?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hm...let me show you what I did: \[0 = 6*(15z)^4*(3z^26)* (2z)  20 * (15z)^3*(3z^26)^2\]\[20 * (15z)^3*(3z^26)^2= 6*(15z)^4*(3z^26)* (2z)\](cancel like terms and move coefficients)\[\frac{5}{3}*(3z^26) = (15z)*z\]\[5z^210=z5z^2\]\[10z^2z10 = 0\] and you can solve after that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ohhhhh...Okay. Thank you very much! I appreciate your help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The first person to answer this was incorrect, right? You should get some weird decimal numbers after using quadratic formula?
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